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jok3333 [9.3K]
3 years ago
13

What is extraneous roots?

Mathematics
1 answer:
kobusy [5.1K]3 years ago
7 0

extraneous root is a solution o an equation that seems to be  right but when we check it (by substituting it into the original equation) it turns out not to be right


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Find the mean and median of the following data set:
VARVARA [1.3K]

Answer:

Mean: 23. 54545455

Median: 22

Step-by-step explanation:

The median represents the center the most because when you are solving for the median, you look for the direct center of the terms which will be 22, the median.

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The cook has a variety of meals to prepare for a villa center’s residents. she averages 16 vegetarian meals every day of the wee
MrRissso [65]

Answer:

110

Step-by-step explanation:

for this problem you need to multiply the daily meals by 7(for each day of the week) getting a grand total of 112 meals per week. since the last digit is under 4, you round down so your final answer is 110

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Find the simple interest for :<br> $668,5%, 2 years
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Step-by-step explanation:

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What is V36x4 in simplest form?
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The answer is A. 6x^2 hope this helps!
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Evaluate the integral by making an appropriate change of variables.
VARVARA [1.3K]

By inspecting the integrand, the "obvious" choice for substitution would be

<em>u</em> = <em>y</em> + <em>x</em>

<em>v</em> = <em>y</em> - <em>x</em>

<em />

Solving for <em>x</em> and <em>y</em>, we would have

<em>x</em> = (<em>u</em> - <em>v</em>)/2

<em>y</em> = (<em>u</em> + <em>v</em>)/2

in which case the Jacobian and its determinant are

J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12

The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.

Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So

-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>

-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2

-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>

-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>

<em />

So, the integral becomes

\displaystyle\iint_R5\cos\left(7\frac{y-x}{y+x}\right)\,\mathrm dA=\int_8^9\int_{-u}^u\frac52\cos\left(\frac{7v}u\right)\,\mathrm dv\,\mathrm du

=\displaystyle\frac52\int_8^9\frac u7(\sin7-\sin(-7))\,\mathrm du

=\displaystyle\frac57\sin7\int_8^9u\,\mathrm du

=\displaystyle\frac5{14}\sin7(9^2-8^2)=\boxed{\frac{85}{14}\sin7}

4 0
2 years ago
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